Problem: If $x$ satisfies $x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$.
Solution: Let $k = x+\frac 1x$. Notice that $k^2 = x^2 + 2 + \frac 1{x^2}$, so $x^2 + \frac 1{x^2} = k^2-2$. Substituting this into the equation gives $(k^2-2) + 3 \cdot (k) = 26$, or $k^2 + 3k - 28 = (k+7)(k-4) = 0$. Since $x$ is positive, then $k > 0$, so $k = 4$. Substituting back, $x + \frac 1x = 4 \Longrightarrow x^2 - 4x + 1 = 0 \Longrightarrow x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$. To match the desired form, we take the solution $x = 2+\sqrt{3}$, and the answer is $\boxed{5}$.